1
Lecture 14: The product and quotient rule
1.1
Outline
• The product rule, the reciprocal rule, and the quotient rule.
• Power rule, derivative the exponential function
• Derivative of a sum
• Differentiability implies continuity.
• Example: Finding a derivative.
1.2
The derivative
Theorem 1 Suppose that f and g are two functions which are differentiable at a
point x, then f g is differentiable at x and
(f g)0 (x) = f 0 (x)g(x) + f (x)g 0 (x).
Proof. The proof depends on rewriting the difference quotient for f g in terms of the
difference quotients for f and g. This depends on the trick of adding and subtracting
f (x)g(x + h) as follows
f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x)
f (x + h)g(x + h) − f (x)g(x)
=
h
h
f (x + h) − f (x)
g(x + h) − g(x)
=
g(x + h) + f (x)
.
h
h
We know that the difference quotients for f and g have a limit as h tends to zero.
Since g is differentiable at x, it is continuous and we have
lim g(x + h) = g(x).
h→0
Thus we may use the rules for sums and products of limits to obtain that
f (x + h) − f (x)
g(x + h) − g(x)
lim g(x + h) + f (x) lim
h→0
h→0
h
h
= f (x)g(x) + f (x)g 0 (x)
(f g)0 (x) = lim
h→0
0
One way to understand this rule is to think of a rectangle whose length ` and
width w are given by `(t) = a + bt and w(t) = c + dt. Then the area will be given by
`(t)w(t) = (`0 + mt)(w0 + nt) = `0 w0 + (mw0 + `0 n)t + mnt2 .
At t = 0, the instantaneous rate of change of the area will be the coefficient of t,
mw0 + `0 n. Since m is the rate of change of ` and n is the rate of change of w, the
rate of change of the product is exactly what we see in the product rule.
Example. Compute that derivative f (x) = x3 ex .
Solution. We use the Leibniz notation,
d 3 x
d
d
(x e ) = ( x3 )ex + x3 ex
dx
dx
dx
= 3x2 ex + x3 ex .
Example. Find the derivative of x6 by writing x6 = x3 · x3 and applying the product
rule.
Solution. We write x6 = x3 · x3 and apply the product rule,
d 3 3
d
d
d 6
x =
(x · x ) = x3 x3 + ( x3 )x3 .
dx
dx
dx
dx
Computing the derivatives gives
3x2 · x3 + x3 · 3x2 = 6x5 .
1.3
Reciprocals
We find the derivative of a reciprocal or the multiplicative inverse of a function.
Theorem 2 If g is differentiable at x and g(x) 6= 0, then 1/g is differentiable at x
and we have
!0
1
−g 0 (x)
(x) =
.
g
g(x)2
Proof. We write out the difference quotient for 1/g, obtain a common denominator
and simplify to express it in terms of the difference quotient for g,
1
1
1
1
g(x)
g(x + h)
(
−
) =
−
h g(x + h) g(x)
h g(x)g(x + h) g(x)g(x + h)
−1
1
=
(g(x + h) − g(x))
.
h
g(x + h)g(x)
Now we may use the limit laws and that 1/g is continuous at x to write
1
g
!0
g(x + h) − g(x)
1
· lim
h→0
h→0 g(x)g(x + h)
h
−g 0 (x)
=
.
g(x)2
(x) = − lim
Example. Use this rule to find the derivative
d 1
.
dx x4
Solution.
d 1
dx x4
=
−4x3
(x4 )2
=
−4
.
x5
We can use the reciprocal rule to extend the power rule to negative exponents.
Example. Use the reciprocal rule to find the derivative
d −n
x ,
dx
Solution.
1.4
d 1
dx xn
=
−nxn−1
x2n
for n = 1, 2, 3, . . . .
= −nx−n−1 .
Quotient rule
Finally we give the quotient rule. Note that it is often simpler to rewrite a quotient
as a product and avoid the quotient rule.
Theorem 3 If f and g are differentiable at x and g(x) 6= 0, then f /g is differentiable
at x and
!0
f
f 0 (x)g(x) − f (x)g 0 (x)
(x) =
.
g
g(x)2
Proof. We may prove this writing f /g = f · g1 and using the product and the reciprocal rule.
!0
f
1
1
−g 0 (x)
(x) = (f )0 (x) = f 0 (x)
+ f (x)
g
g
g(x)
g(x)2
We may simplify this last expression by obtaining a common denominator.
f 0 (x)
−g 0 (x)
g(x)
−g 0 (x)
f 0 (x)g(x) − f (x)g 0 (x)
1
0
+ f (x)
=
f
(x)
+
f
(x)
=
.
g(x)
g(x)2
g(x)2
g(x)2
g(x)2
Example. Find the tangent line to
f (x) =
x2 − 1
x2 + 1
at x = 1.
Solution. The tangent line will pass through the point (1, f (1)) = (1, 0). We need the
derivative of f to compute the slope. We use the quotient rule to find the derivative
of f ,
d
d
(x2 − 1))(x2 + 1) − (x2 − 1) dx
(x2 + 1)
( dx
(x2 + 1)2
2x(x2 + 1) − (x2 − 1)2x
=
(x2 + 1)2
4x
.
=
(x2 + 1)2
f 0 (x) =
At 1, we have f 0 (1) = 1. Thus the tangent line has the equation
y − 0 = 1(x − 1).
We simplify this to give y = x − 1 as the equation of the tangent line.
September 25, 2012